import java.util.*;

/**
 * @author LKQ
 * @date 2022/5/15 16:16
 * @description 动态规划
 */
public class Solution {
    public static void main(String[] args) {
        Solution solution = new Solution();
        solution.isMatch("abcdef", "a*");
    }
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        // dp[i][j]表示 s[0..i] 与 p[0..j]子字符串能够匹配
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i <= n; ++i) {
            if (p.charAt(i - 1) == '*') {
                // p中如果是连续的*，那么到连续*之前，都可以匹配空字符串
                dp[0][i] = true;
            } else {
                break;
            }
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (p.charAt(j - 1) == '*') {
                    // dp[i][j-1]为不使用*号，能否匹配，dp[i-1][j]为使用*号，

                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                } else if (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }
        return dp[m][n];
    }
}
